Question

A block of mass M = 5.80 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 6250 N/m. A bullet of mass m = 8.30 g and velocity of magnitude 570 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

Answer 1

Part A.

Since bullet is embedded in the block, SO this is inelastic collision

Using momentum conservation

Pi = Pf

mb*Vb + Mb*Vb = (mb + Mb)*V

mb = mass of bullet = 8.30 gm

Mb = mass of block = 5.80 kg

vb = initial speed of bullet = 570 m/sec

Vb = initial speed of Block = 0 m/sec

V = final speed of block + bullet = ?

V = (mb*vb + Mb*Vb)/(mb + Mb)

V = (8.30*10^-3*570 + 5.80*0)/(8.30*10^-3 + 5.80)

V = 0.814 m/sec

Part B.

Now Using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at equilibrium

KEf = 0, at max amplitude

KEi = initial KE of bullet + block = 0.5*(mb + Mb)*V^2

PEf = 0.5*k*A^2

Where, A = amplitude of SHM

So,

0.5*(mb + Mb)*V^2 = 0.5*k*A^2

A = V*sqrt ((mb + Mb)/k)

A = 0.814*sqrt ((8.30*10^-3 + 5.80)/6250)

A = 0.0248 m = 2.48 cm

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